Question 770548
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There are 10 possible digits, 0 through 9, for each place in the code.  But if a particular place must be either even or odd, then there are only 5 choices for that place -- because of the 10 possible digits, 5 of them are even and 5 of them are odd.


In the case of the old code, there are 5 ways to choose the first digit: it can either be 1, 3, 5, 7, or 9.  Then, for each of those choices, there are 5 ways to pick the second digit because you are allowed to repeat digits, namely 1, 3, 5, 7, and 9 again.  So just considering the first two positions, there are 5 times 5, which is to say 25 ways to fill them.  Then, the next position has 5 ways to fill it...0, 2, 4, 6, 8.  And so on.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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